Question
The function \(f\) is defined by \(f(x) = \frac{{{{\rm{e}}^x} + {{\rm{e}}^{ – x}}}}{2}\) .
(i) Obtain an expression for \({f^{(n)}}(x)\) , the nth derivative of \(f(x)\) with respect to \(x\).
(ii) Hence derive the Maclaurin series for \(f(x)\) up to and including the term in \({x^4}\) .
(iii) Use your result to find a rational approximation to \(f\left( {\frac{1}{2}} \right)\) .
(iv) Use the Lagrange error term to determine an upper bound to the error in this approximation.
Use the integral test to determine whether the series \(\sum\limits_{n = 1}^\infty {\frac{{\ln n}}{{{n^2}}}} \) is convergent or divergent.
Answer/Explanation
Markscheme
(i) \({f^{(n)}}(x) = \frac{{{{\rm{e}}^x} + {{( – 1)}^n}{{\rm{e}}^{ – x}}}}{2}\) (M1)A1
(ii) Coefficient of \({x^n} = \frac{{{f^{(n)}}(0)}}{{n!}}\) (M1)
\( = \frac{{1 + {{( – 1)}^n}}}{{2n!}}\) (A1)
\(f(x) = 1 + \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{24}} + \ldots \) A1
(iii) Putting \(x = \frac{1}{2}\) M1
\(f(0.5) = 1 + \frac{1}{8} + \frac{1}{{16 \times 24}} = \frac{{433}}{{384}}\) (M1)A1
(iv) Lagrange error term \( = \frac{{{f^{(n + 1)}}(c)}}{{(n + 1)!}}{x^{n + 1}}\) M1
\( = \frac{{{f^{(5)}}(c)}}{{120}} \times {\left( {\frac{1}{2}} \right)^5}\) A1
\({{f^{(5)}}(c)}\) is an increasing function because – any valid reason, e.g. plotted a graph, positive derivative, increasing function minus a decreasing function, so this is maximized when \(x = 0.5\) . R1
Therefore upper bound \( = \frac{{({{\rm{e}}^{0.5}} – {{\rm{e}}^{ – 0.5}})}}{{2 \times 120}} \times {\left( {\frac{1}{2}} \right)^5}\) M1
\( = 0.000136\) A1
[13 marks]
We consider \(\int_1^\infty {\frac{{\ln x}}{{{x^2}}}} {\rm{d}}x = \int_1^\infty {\ln x{\rm{d}}x} \left( { – \frac{1}{2}} \right)\) M1A1
\( = \left[ { – \frac{{\ln x}}{x}} \right]_1^\infty + \int_1^\infty {\frac{{1x}}{{{x^2}}}} {\rm{d}}x\) A1A1
\( = \left[ { – \frac{{\ln x}}{x}} \right]_1^\infty – \left[ {\frac{1}{x}} \right]_1^\infty \) A1
Now \(\mathop {\lim }\limits_{x \to \infty } \left( {\frac{1}{x}} \right) = 0\) R1
\(\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{\ln x}}{x}} \right) = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{1}{x}} \right) = 0\) M1A1
The integral is convergent with value \(1\) and so therefore is the series. R1
[9 marks]
Question
Let \({S_n} = \sum\limits_{k = 1}^n {\frac{1}{k}} \) .
Show that, for \(n \ge 2\) , \({S_{2n}} > {S_n} + \frac{1}{2}\) .
Deduce that \({S_{2m + 1}} > {S_2} + \frac{m}{2}\) .
Hence show that the sequence \(\left\{ {{S_n}} \right\}\) is divergent.
Answer/Explanation
Markscheme
\({S_{2n}} = {S_n} + \frac{1}{{n + 1}} + \frac{1}{{n + 2}} + \ldots + \frac{1}{{2n}}\) M1
\( > {S_n} + \frac{1}{{2n}} + \frac{1}{{2n}} + \ldots + \frac{1}{{2n}}\) M1A1
\( = {S_n} + \frac{1}{2}\) AG
[3 marks]
Replacing \(n\) by \(2n\),
\({S_{4n}} > {S_{2n}} + \frac{1}{2}\) M1A1
\( > {S_n} + 1\) A1
Continuing this process,
\({S_{8n}} > {S_n} + \frac{3}{2}\) (A1)
In general,
\({S_{{2^m}n}} > {S_n} + \frac{m}{2}\) M1A1
Putting \(n = 2\) M1
\({S_{{2^{m + 1}}}} > {S_2} + \frac{m}{2}\) AG
[7 marks]
Consider the (large) number \(N\). M1
Then, \({S_{2m + 1}} > N\) if \({S_2} + \frac{m}{2} > N\) A1
i.e. if \(m > 2(N – {S_2})\) A1
This establishes the divergence. AG
[3 marks]
Question
The function \(f(x)\) is defined by the series \(f(x) = 1 + \frac{{(x + 2)}}{{3 \times 1}} + \frac{{{{(x + 2)}^2}}}{{{3^2} \times 2}} + \frac{{{{(x + 2)}^3}}}{{{3^3} \times 3}} + \ldots \) .
Write down the general term.
Find the interval of convergence.
Solve the differential equation \((u + 3{v^3})\frac{{{\rm{d}}v}}{{{\rm{d}}u}} = 2v\) , giving your answer in the form \(u = f(v)\) .
Answer/Explanation
Markscheme
the general term is \(\frac{{{{(x + 2)}^n}}}{{{3^n}n}}\) A1
[1 mark]
\(\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right) = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{{{(x + 2)}^{n + 1}}}}{{{3^{n + 1}}(n + 1)}} \times \frac{{{3^n}n}}{{{{(x + 2)}^n}}}} \right]\) M1A1A1
\( = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{(x + 2){n^{}}}}{{3(n + 1)}}} \right]\) A1
\( = \frac{{(x + 2)}}{3}\) since \( = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{{n^{}}}}{{n + 1}}} \right] = 1\) A1R1
the series is convergent if \(\left| {\frac{{(x + 2)}}{3}} \right| < 1\) R1
then \( – 3 < x + 2 < 3 \Rightarrow – 5 < x < 1\) A1
if \(x = – 5\) , series is \(1 – 1 + \frac{1}{2} – \frac{1}{3} + \ldots + \frac{{{{( – 1)}^n}}}{n} + \ldots \) which converges M1A1
if \(x = 1\) , series is \(1 + 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n} + \ldots \) which diverges M1A1
the interval of convergence is \( – 5 \le x < 1\) A1
[13 marks]
\((u + 3{v^3})\frac{{{\rm{d}}v}}{{{\rm{d}}u}} = 2v\)
\(\frac{{{\rm{d}}u}}{{{\rm{d}}v}} = \frac{{(u + 3{v^3})}}{{2v}} = \frac{u}{{2v}} + \frac{{3{v^2}}}{2}\) M1A1
\(\frac{{{\rm{d}}u}}{{{\rm{d}}v}} – \frac{u}{{2v}} = \frac{{3{v^2}}}{2}\) A1
IF is \({{\rm{e}}^{\int {\frac{1}{{2v}}} {\rm{d}}v}} = {{\rm{e}}^{\frac{1}{2}\ln v}}\) M1
\( = {v^{\frac{1}{2}}}\) A1
\(\frac{u}{{\sqrt v }} = \int {\frac{{3{v^{\frac{3}{2}}}}}{2}} {\rm{d}}v\) M1
\( = \frac{3}{5}{v^{\frac{5}{2}}} + c\) A1
\(u = \frac{3}{5}{v^3} + c\sqrt v \) A1
[8 marks]
Question
The diagram shows a sketch of the graph of \(y = {x^{ – 4}}\) for \(x > 0\) .
By considering this sketch, show that, for \(n \in {\mathbb{Z}^ + }\) ,\[\sum\limits_{r = n + 1}^\infty {\frac{1}{{{r^4}}}} < \int_n^\infty {\frac{{{\rm{d}}x}}{{{x^4}}}} < \sum\limits_{r = n}^\infty {\frac{1}{{{r^4}}}} .\]
Let \(S = \sum\limits_{r = 1}^\infty {\frac{1}{{{r^4}}}} \) .
Use the result in (a) to show that, for \(n \ge 2\) , the value of \(S\) lies between
\(\sum\limits_{r = 1}^{n – 1} {\frac{1}{{{r^4}}}} + \frac{1}{{3{n^3}}}\) and \(\sum\limits_{r = 1}^n {\frac{1}{{{r^4}}}} + \frac{1}{{3{n^3}}}\) .
(i) Show that, by taking \(n = 8\) , the value of \(S\) can be deduced correct to three decimal places and state this value.
(ii) The exact value of \(S\) is known to be \(\frac{{{\pi ^4}}}{N}\)where \(N \in {\mathbb{Z}^ + }\) . Determine the value of \(N\) .
Now let \(T = \sum\limits_{r = 1}^\infty {\frac{{{{( – 1)}^{r + 1}}}}{{{r^4}}}} \) .
Find the value of \(T\) correct to three decimal places.
Answer/Explanation
Markscheme
(M1)
total area of “upper” rectangles
\( = \frac{1}{{{n^4}}} \times 1 + \frac{1}{{{{(n + 1)}^4}}} \times 1 + \frac{1}{{{{(n + 2)}^4}}} \times 1 + \ldots = \sum\limits_{r = n}^\infty {\frac{1}{{{r^4}}}} \) M1A1
total area of “lower” rectangles
\( = \frac{1}{{{{(n + 1)}^4}}} \times 1 + \frac{1}{{{{(n + 2)}^4}}} \times 1 + \frac{1}{{{{(n + 3)}^4}}} \times 1 + \ldots = \sum\limits_{r = n + 1}^\infty {\frac{1}{{{r^4}}}} \) A1
the total area under the curve from \(x = n\) to infinity lies between these two sums hence \(\sum\limits_{r = n + 1}^\infty {\frac{1}{{{r^4}}}} < \int_n^\infty {\frac{{{\rm{d}}x}}{{{x^4}}}} < \sum\limits_{r = n}^\infty {\frac{1}{{{r^4}}}} \) R1AG
[5 marks]
first evaluate the integral
\(\int_n^\infty {\frac{{{\rm{d}}x}}{{{x^4}}}} = – \left[ {\frac{1}{{3{x^3}}}} \right]_n^\infty = \frac{1}{{3{n^3}}}\) M1A1
it follows that
\(\sum\limits_{r = n + 1}^\infty {\frac{1}{{{r^4}}}} < \frac{1}{{3{n^3}}}\) A1
adding \(\sum\limits_{r = 1}^n {\frac{1}{{{r^4}}}} \) to both sides, M1
\(S < \sum\limits_{r = 1}^n {\frac{1}{{{r^4}}}} + \frac{1}{{3{n^3}}}\) A1
similarly,
\(\sum\limits_{r = n}^\infty {\frac{1}{{{r^4}}}} > \frac{1}{{3{n^3}}}\) A1
adding \(\sum\limits_{r = 1}^{n – 1} {\frac{1}{{{r^4}}}} \) to both sides, M1
\(S > \sum\limits_{r = 1}^{n – 1} {\frac{1}{{{r^4}}}} + \frac{1}{{3{n^3}}}\) A1
hence the value of \(S\) lies between
\(\sum\limits_{r = 1}^{n – 1} {\frac{1}{{{r^4}}}} + \frac{1}{{3{n^3}}}\) and \(\sum\limits_{r = 1}^n {\frac{1}{{{r^4}}}} + \frac{1}{{3{n^3}}}\) AG
[8 marks]
(i) putting \(n = 8\) , we find that
\(S < 1.08243 \ldots \) and \(S > 1.08219 \ldots \) A1A1
it follows that \(S = 1.082\) to 3 decimal places A1
(ii) substituting this value of \(S\),
\(N \approx \frac{{{\pi ^4}}}{{1.082}} \approx 90.0268\) M1A1
\(N = 90\) A1
[6 marks]
EITHER
successive partial sums are
1 M1
0.9375
0.9498…
0.9459…
0.9475…
0.9467…
0.9471… A1
it follows that correct to 3 decimal places \(T = 0.947\) A1
OR
\(T = S – \frac{2}{{16}}S\) M1A1
using part (c)(i) or \(0.94703…\) using the sum given in part (c)(ii) \(0.9471…\)
it follows that \(T = 0.947\) correct to 3 decimal places A1
[3 marks]
Question
Find the value of \(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} – \cot x} \right)\) .
Find the interval of convergence of the infinite series\[\frac{{(x + 2)}}{{3 \times 1}} + \frac{{{{(x + 2)}^2}}}{{{3^2} \times 2}} + \frac{{{{(x + 2)}^3}}}{{{3^3} \times 3}} + \ldots \]
(i) Find the Maclaurin series for \(\ln (1 + \sin x)\) up to and including the term in \({x^3}\) .
(ii) Hence find a series for \(\ln (1 – \sin x)\) up to and including the term in \({x^3}\) .
(iii) Deduce, by considering the difference of the two series, that \(\ln 3 \simeq \frac{\pi }{3}\left( {1 + \frac{{{\pi ^2}}}{{216}}} \right)\) .
Answer/Explanation
Markscheme
EITHER
\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} – \cot x} \right)\)
\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\tan x – x}}{{x\tan x}}} \right)\) M1A1
\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{{{\sec }^2}x – 1}}{{x{{\sec }^2}x + \tan x}}} \right)\) , using l’Hopital A1
\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{2{{\sec }^2}x\tan x}}{{2{{\sec }^2}x + 2x{{\sec }^2}x\tan x}}} \right)\) A1A1
\( = 0\) A1
OR
\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} – \cot x} \right)\)
\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x – x\cos x}}{{x\sin x}}} \right)\) M1A1
\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{x\sin x}}{{\sin x + x\cos x}}} \right)\) , using l’Hopital A1
\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x + x\cos x}}{{2\cos x – x\sin x}}} \right)\) A1A1
\( = 0\) A1
[6 marks]
\({u_n} = \frac{{{{(x + 2)}^n}}}{{{3^n} \times n}}\) A1
\(\frac{{{u_{n + 1}}}}{{{u_n}}} = \frac{{\frac{{{{(x + 2)}^{n + 1}}}}{{{3^{n + 1}} \times (n + 1)}}}}{{\frac{{{{(x + n)}^n}}}{{{3^n} \times n}}}} = \frac{{(x + 2)n}}{{3(n + 1)}}\) M1A1
\(\mathop {\lim }\limits_{n \to \infty } \frac{{(x + 2)n}}{{3(n + 1)}} = \frac{{(x + 2)}}{3}\) M1A1
\(\left| {\frac{{(x + 2)}}{3}} \right| < 1 \Rightarrow – 5 < x < 1\) M1A1
if \(x = 1\) series is \(1 + \frac{1}{2} + \frac{1}{3} + \ldots \) which diverges A1
if \(x = – 5\) series is \( – 1 + \frac{1}{2} – \frac{1}{3} + \ldots + \frac{{{{( – 1)}^n}}}{n}\) which converges A1
hence interval is \( – 5 \le x < 1\) A1
[10 marks]
(i) \(f(x) = \ln (1 + \sin x)\) , \(f(0) = 0\) A1
\(f'(x) = \frac{{\cos x}}{{1 + \sin x}}\) , \(f'(0) = 1\) A1
\(f”(x) = \frac{{ – \sin x(1 + \sin x) – {{\cos }^2}x}}{{{{(1 + \sin x)}^2}}} = \frac{{ – (1 + \sin x)}}{{{{(1 + \sin x)}^2}}} = \frac{{ – 1}}{{1 + \sin x}}\) , \(f”(0) = – 1\) A1
\(f”'(x) = \frac{{\cos x}}{{{{(1 + \sin x)}^2}}}\) , \(f”'(0) = 1\) A1
\(\ln (1 + \sin x) \approx x – \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} – \ldots \) A1
(ii) \( – \sin x = \sin ( – x)\) M1
so, \(\ln (1 – \sin x) \approx – x – \frac{{{x^2}}}{2} – \frac{{{x^3}}}{6} – \ldots \) A1
(iii) \(\ln (1 + \sin x) – ln(1 – \sin x)\)
\( = \ln \left( {\frac{{1 + \sin x}}{{1 – \sin x}}} \right) \approx 2x + \frac{{{x^3}}}{3}\) M1A1
let \(x = \frac{\pi }{6}\) then, \(\ln \left( {\frac{{1 + \frac{1}{2}}}{{1 – \frac{1}{2}}}} \right) = \ln 3 \approx 2\left( {\frac{\pi }{6}} \right) + \frac{{{{\left( {\frac{\pi }{6}} \right)}^3}}}{3}\) M1A1A1
\( = \frac{\pi }{3}\left( {1 + \frac{{{\pi ^2}}}{{216}}} \right)\) AG
[12 marks]
Question
(i) Show that the improper integral \(\int_0^\infty {\frac{1}{{{x^2} + 1}}} {\rm{d}}x\) is convergent.
(ii) Use the integral test to deduce that the series \(\sum\limits_{n = 0}^\infty {\frac{1}{{{n^2} + 1}}} \) is convergent, giving reasons why this test can be applied.
(i) Show that the series \(\sum\limits_{n = 0}^\infty {\frac{{{{( – 1)}^n}}}{{{n^2} + 1}}} \) is convergent.
(ii) If the sum of the above series is \(S\), show that \(\frac{3}{5} < S < \frac{2}{3}\) .
For the series \(\sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{{n^2} + 1}}} \)
(i) determine the radius of convergence;
(ii) determine the interval of convergence using your answers to (b) and (c).
Answer/Explanation
Markscheme
(i) consider \(\int_0^R {\frac{1}{{{x^2} + 1}}} {\rm{d}}x\) M1
\( = \left[ {\arctan (x)} \right]_0^R = \arctan (R)\) A1
\(\mathop {\lim }\limits_{R \to \infty } \arctan (R) = \frac{\pi }{2}\) (a finite number) R1
hence the improper integral is convergent AG
(ii) the terms of the series are positive A1
the terms are decreasing A1
the terms tend to zero A1
by the integral test, the series converges AG
[6 marks]
(i) the absolute values of the terms are monotonically decreasing A1
to zero A1
the series converges by the alternating series test R1AG
Note: Accept absolute convergence, with reference to part (b)(ii) \( \Rightarrow \) convergence.
(ii) statement that successive partial sums bound the total sum R1
\(S > \frac{1}{1} – \frac{1}{2} + \frac{1}{5} – \frac{1}{{10}} = \frac{3}{5}\) A1
\(S < \frac{1}{1} – \frac{1}{2} + \frac{1}{5} – \frac{1}{{10}} + \frac{1}{{17}} = 0.6588\) A1
\(S < 0.6588 < \frac{2}{3}\) AG
[6 marks]
(i) consider \(\left| {\frac{{\frac{{{x^{n + 1}}}}{{{{(n + 1)}^2} + 1}}}}{{\frac{{{x^n}}}{{{n^2} + 1}}}}} \right|\) M1
\( = \left| {\frac{{x({n^2} + 1)}}{{{{(n + 1)}^2} + 1}}} \right|\) A1
\( \to \left| x \right|\) as \(n \to \infty \) A1
therefore radius of convergence \( = 1\) A1
(ii) interval of convergence \( = \left[ { – 1,1} \right]\) A1A1
Note: A1 for [\( – 1\), and A1 for \(1\)].
[6 marks]